Equation of the plane containing the following point and line?

Question

I've been trying to find the equation for the plane containing a specific point and line, however, I can't seem to get the correct answer.

The point and line in question are $(2,1,1)$ and the line A given by $3a + b - c = 2$, $-2b + c = 1$.

I began by finding the parametric form of the line given, resulting in the following: $(1-s,-3s,1-6s)$. Hence the direction vector is $(-1,-3,-6)$ and then using the given point $(2,1,1)$ we can get a second direction vector for the plane: $(3,4,7)$. Taking the cross product of these two direction vectors gives us the normal to the plane: $(3,-11,5)$ which we can then dot product with $(a - 2, b - 1, c - 1)$ and set equal to zero to find the equation of the plane, which turns out as $3a - 11b + 5c = 0$.

This equation does not seem to be correct when testing points on the line, for example, $(0,-3,-5)$. Can anyone spot the error in what I have done?

Thanks a lot!

Lauren

Answer 1

Hint. All the planes which contain the line $A$ have the form $$\lambda(3x+y-z-2)+\mu(-2y+z-1)=0$$ for $\lambda,\mu\in\mathbb{R}$ (a linear combination of the two given planes).

Now the plane which passes through $(2,1,1)$ should satisfy $$\lambda(6+1-1-2)+\mu(-2+1-1)=0\Rightarrow 4\lambda=2\mu.$$
Hence, by letting $\mu=2$ and $\lambda=1$, we get the required plane $$3x-3y+z=4.$$

Answer 2

Your line is correct.

$$(1−s,−3s,1−6s)=(1,0,1)+s(-1,-3,-6)$$

Second direction vector of your plane is wrong though. it should be

$$(2,1,1)-(1,0,1)=(1,1,0)$$

(why?)

can you solve it from here?

Answer 3

Following your way I get the following parametrisation

$$ \left\{ \begin{array}{l} x=s \\ y=3s-3 \\ z=6s-5 \end{array} \right. $$

So a direction vector is $\mathbf{u}=(1,3,6)^T$ (proportional to the one you found). Another direction vector is given by the two points $(2,1,1)$ and the point of parameter $s=0$ i.e. $(0,-3,-5)$ and this gives $\mathbf{v}=(2,4,6)^T$. Now a normal vector to the plane is given by

$$\mathbf{u}\times{\mathbf{v}\over 2}=\begin{pmatrix}-3&\\3&\\-1&\end{pmatrix}$$

And this in turn gives the following equation for the plane

$$3x-3y+z+d=0$$

And the constant term $d$ is such that $3\times 2-3\times 1+1+d=0$ so $d=-4$ (because $(2,1,1)\in\text{ plane}$) and the plane has the following equation

$$3x-3y+z-4=0$$